physics ncert solutions class 11th

Kindly consider the following figure

From conservation of Angular momentum about hinged point

$mv\mathcal{l}=\left(\frac{m{\mathcal{l}}^2}{3}+2m.{\mathcal{l}}^2\right)\omega$            

$\omega =\left(\frac{3}{7}\frac{v}{\mathcal{l}}\right)$            

Now from conservation of Energy.

$\frac{1}{2}*\frac{7}{3}m{\mathcal{l}}^2*{\omega }^2=mgl+2mg.2\mathcal{l}$            

$v=\sqrt[]{\frac{70}{3}g\mathcal{l}}$            

Since plane is smooth hence motion is only translational

$a=gsin\theta =10*\frac{1}{2}=5m/s^2$              

$s=ut+\frac{1}{2}a{t}^2$            

$1.6=0+\frac{1}{2}*5*t^2$            

$t^2=\sqrt[]{\frac{3.2}{5}}=\sqrt[]{0.64}=0.8sec.$            

In isothermal process, temperature is constant.

In isochoric process, volume is constant.

In adiabatic process, there is no exchange of heat.

In isobaric process, pressure is constant.

$V_{rms}=\sqrt[]{\frac{3RT}{M}}$

& $v_P=\sqrt[]{\frac{2RT}{M}}$

$v_{rms}=\sqrt[]{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{v}_P$

λ? = h/√2mE? = λ? = h/√2mE?
=> E? = (4/9)E? = 4eV
E? = E? - eV? => V? = 5V

$\mathrm{T}\mathrm{i}=-{50}^{?}\mathrm{C}$

$\begin{array}{rr}& =223\mathrm{K}\\ & {\mathrm{V}}_{\mathrm{r}\mathrm{m}\mathrm{s}}\propto \sqrt[]{\mathrm{T}}\end{array}$

As $\begin{array}{rr}& =223\mathrm{K}\\ & {\mathrm{V}}_{\mathrm{r}\mathrm{m}\mathrm{s}}\propto \sqrt[]{\mathrm{T}}\end{array}$  increased by 3 times

So ${\left({\mathrm{v}}_{\mathrm{r}\mathrm{m}\mathrm{s}}\right)}_{\mathrm{f}}=4{\left({\mathrm{v}}_{\mathrm{r}\mathrm{m}\mathrm{s}}\right)}_{\text{initial}}$

${\mathrm{T}}_{\mathrm{f}}=16{\mathrm{T}}_{\mathrm{i}}$

$=3568\mathrm{K}$

${\mathrm{T}}_{\mathrm{f}}=(3568-273{)}^{?}\mathrm{C}$

$={3295}^{?}\mathrm{C}$

Root mean square speed of gas molecules

${\nu }_{\text{rms}}=\sqrt[]{\frac{3RT}{M}}$

Pressure exerted by ideal Gas

$\mathrm{P}=\frac{1}{3}\rho v_{\mathrm{r}\mathrm{m}\mathrm{s}}^2$

$\mathrm{P}=\frac{1}{3}\mathrm{m}\mathrm{n}{\nu }^2$

$\rho =\mathrm{m}\mathrm{n},{\mathrm{v}}_{\mathrm{r}\mathrm{m}\mathrm{s}}^2={\nu }^{-2}$

Average kinetic energy of a molecular

$\mathrm{K}\mathrm{E}=\frac{3}{2}\mathrm{K}\mathrm{T}$

Total internal energy of 1 mole of a diatomic gas

$\mathrm{U}=\frac{\mathrm{f}}{2}\mu \mathrm{R}\mathrm{T}$

$\mathrm{U}=\frac{5}{2}\mathrm{R}\mathrm{T}$ (For 1 mole diatomic gas)

(n) $\left(1.1\right)=(\mathrm{n}+1)$

$0.1\left(\mathrm{n}\right)=1$

$\mathrm{n}=10$

Required number of oscillation $=n+1=10+1=11$